A Note on the Group Inverses of Block Matrices Over Rings

Transcription

1 Filomat 31:1 017, Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: A Note on the Group Inverses of Block Matrices Over Rings Hanyu Zhang a a Group of Mathematical, Jidong second middle school, Jidong 15800, PR China Abstract Suppose R is an associative ring with identity 1 The purpose of this paper is to give some necessary and sufficient conditions for the existence and the representations of the group inverse of the block matrix and M under some conditions Some examples are given to illustrate our results 1 The first section The Drazin group inverse of block matrices have numerous applications in many areas, especially in singular differential and difference equations and finite Markov chains see[1,4,5,7,15] In 1979, Campbell and Meyer proposed a problem to find a concrete expression for Drazin group inverse of block matrices M, where A and D are square but need not to be the same sizesee[1] Although the problem has not been solved completely yet, ones have got results on group inverses under some special conditions in [,3,6,9-11,16-18,1] The purpose of this paper is to extend some recent results on the group inverse of block matrices, as for basic ring and matrix type In [3], Li et alinvestigate the group inverse of the block matrix over skew filed under the conditions A exists, XA AX, and ra rax This generalizes Theorem 11 of [8] In this paper, the sufficient and necessary condition for the existence of the group inverse of the above partitioned matrix over any ring is characterized In [13], Bu et alinvestigated the group inverse of the block matrix over skew filed under the conditions A is invertible and D CA 1 B exists In [19], Deng et alstudied the group inverse of the block matrix over complex Banach spaces under the conditions A and S D CA B are group invertible, A π B 0 and S π C 0 In this paper, we will investigate the group inverse of the above block matrix over rings under weaker conditions We should pointed that the methods of solving problems are quite different from those of [13, 14, 19, 0] 010 Mathematics Subject Classification Primary 15A09; Secondary 16U30 Keywords Ring; Group inverse; Block matrix; anti-triangular Received: 04 March 015; Accepted: 11 July 015 Communicated by Dragan S Djordjević address: zhanghanyu43@16com Hanyu Zhang

2 H Zhang / Filomat 31:1 017, Let R, K be an associative ring with identity 1 and skew field, respectively R m n be the set of all m n matrices over R We denote R m 1 and R 1 n by R m and R n, respectively For A R n n, if there exists a matrix X R n n such that AXA A, XAX X and AX XA, then X is called the group inverse of A and can be denoted by A For A R n m, if X satisfies only AXA A, then A is called regular and X is called a {1}-inverse or regular inverse of A In this case, denote the set of all {1}-inverse of A by A{1} Let A 1 be any {1}-inverse of A, denote I A 1 A and I AA 1 by A πl and A πr respectively If A K m n, ra denotes the rank of A If A is group invertible, we denote I AA by A π, where I is an identity matrix of order n In this paper, for A R m n, we also denote by RA {Ax x R n } and R r A {xa x R m } the range and the row range of A, respectively Some Lemmas In the next section we will use the following results Lemma 1 [] Let A R n n, the followings are equivalent: i A exists; ii A A X and A YA for some X, Y R n n In this case, A YAX AX Y A; iii RA RA, R r A R r A Lemma Let A, X R n n If AX XA, A exists, then A X XA Proof The proof of this Lemma is similar to that of Lemma 31 in [4], so we omit it here Lemma 3 Let A, X R n n If AX XA, A exists, RA RAX and R r A R r XA, then XAA exists and the following equalities hold i XA XAA A ; ii XAA A X A ; iii XAA AA XAA ; iv AA XAA XAA Proof Since RA RAX, there exists a matrix Y over R such that A AXY Using Lemma, we have XAA XA A XA AXY XAA XAA Y, ie, RXAA R[XAA ], so RXAA R[XAA ] Similarly, using R r A R r XA we also get R r XAA R r [XAA ] By Lemma 1, XAA exists i Notice that R r A R r XA, then there exists a matrix Z R n n such that A ZXA Hence XA XAA A 4 X[A ] XAA ZXA 4 X[A ] XAA ZA [XAA ] XAA ZA XAA ZXA ZXAA A ii Similarly, using RA RAX we can obtain that XAA A X A iii By i, we have XAA AA XAA [A ] A XAA [A ] XA XAA XAA XAA XAA XAA iv Similarly, from ii, we can prove iv Lemma 4 Let A R m n, B R n m, RBAB RB and R r BAB R r B, then AB and BA exist and the following equalities hold i AB A[BA ] B; ii BA B[AB ] A; iii AB A ABA ; iv BA B BAB Proof The proof of this Lemma is similar to those of Lemma and 3 in [1]

3 H Zhang / Filomat 31:1 017, Main Results We begin with the following theorem Theorem 31 Let M, where A, B, X, Y R n n, A exists, XA AX, RA RAX and R r A R r XA, then we have i M exists if and only if RB RBSB, R r B R r BSB, where S A π Y XAA ; ii If M exists, then M M1 M, where M 3 M 4 M 1 SB A π + SB π A XAA SB π ZBSB A π, M SB XAA + SB π A [XAA ] + SB π ZBSB [XAA ], M 3 BSB A XAA + B[SB ] A π + BSB ZBSB A π, M 4 BSB A [XAA ] B[SB ] XAA BSB ZBSB XAA, Z A [XAA ] + A XAA Y Proof i The only if part Note that YB A I 0 M X I, 1 Using Lemma 3, we know that XAA exists, and YBSB XA M + YBA BSB BA M I Y 0 I I 0 A YB I I 0 A XAA B I I 0, X I XAYB + AB XA I 0 3 BYB BA X I By Lemma 1, there exist matrices X and Y over R such that M M X and YM M Let I 0 I 0 I 0 X1 X X I 0 X I A XAA B I A, 4 YB I X 3 X 4 X I Y Y1 Y Y 3 Y 4 I Y 0 I From Eqs 1,, 4 and M M X, we can obtain following equations YBSBX 1 + XA + YBAX 3 YB, 6 YBSBX + XA + YBAX 4 A, 7 5 BSBX 1 + BAX 3 B, BSBX + BAX It follows, from Eqs 1, 3, 5 and YM M, that we have Y 1 XAYB + AB + Y BYB YB, 10

4 Y 1 XA + Y BA A, H Zhang / Filomat 31:1 017, Y 3 XAYB + AB + Y 4 BYB B, 1 Y 3 XA + Y 4 BA 0 Instead of 8, we have YBSBX 1 + YBAX 3 YB Substituting it into 6, we have XA X 3 0, so XAA A XX 3 0 By Lemma 3 ii, it follows by A X 3 0, so AX 3 0 Substituting AX 3 0 into 8, we get BSBX 1 B, ie, RBSB RB From 13, by Lemma 3 i, we can get and Y 4 BAA YB Y 3 XA A YB Y 3 XAYB, 14 Y 4 BAA XAA B Y 3 XA A XAA B Y 3 AB 15 Substitute 14 into 1, we have Y 4 BAA YB + Y 3 AB + Y 4 BYB B 16 Substitute 15 into 16, by Lemma 3 iv, we have Y 4 BAA YB Y 4 BAA XAA B + Y 4 BYB B, that is Y 4 BSB B, which implies R r B R r BSB This completes proof of only if part In what follows, we give the proof of if part It follows from Lemma 4 that RBSB RB and R r BSB R r B imply SB and BS exist Let and X 1 SB, X SB XAA, X 3 0, X 4 A XAA Y 1 SB π A XAA, Y SB S, Y 3 BSB A XAA, Y 4 BS We can easily obtain that RBSB RB implies BSBSB B We claim that X 1, X, X 3 and X 4 satisfy the Eqs 6-9 Next, we verify the claim by computation separately 1 YBSBX 1 + XA + YBAX 3 YBSBSB YB; YBSBX + XA + YBAX 4 YBSBSB AA XAA + XA + YBAA XAA YBAA XAA + XAXAA + YBAA XAA A; 3 BSBX 1 + BAX 3 BSBSB B; 4 BSBX + BAX 4 BSBSB AA XAA + BAA XAA BAA XAA + BAA XAA 0 From S A π Y XAA, we have SB YB AA YB AA XAA B By Lemma 4, we know BSB BS B By Lemma 3 and computations, we also can verify Y 1, Y, Y 3 and Y 4 are the solutions of This shows that there exist matrices X and Y over R such that M M X YM hold Hence, by Lemma 1, M exists ii By Lemma 1 and Lemma 3, the expression of M can be obtained from M YMX 11 13

5 H Zhang / Filomat 31:1 017, Example for Theorem 31: Let R Z/6, M, where 1 1 A, B, X, Y AX, YB 0 4 It is easy to verity AX XA, RA RAX and R r A R r XA By a direct computation, we know that XAA and SB exist Further, we have A, A π, XAA 1, SB 0 SB 0 4, SB π , Z 0 0 Clearly, RB RBSB and R r B R r BSB By Theorem 1, M exists and 4 M The following corollary follow Theorem 1 Corollary 3 [3, Theorem1], Let M, where A, B, X, Y K n n, A exists, XA AX, ra rax, then i M exists if and only if rb rbsb, where S A π Y XAA ; ii If M exists, then the representation of M is the same as in Theorem 1 Proof When R K, it is easy to see that Whence the corollary is easily proved ra rax RA RAX and R r A R r XA; rb rbsb RB RBSB and R r B R r BSB Next, we consider the generalizations of some results in [13, 14, 19, 0] Theorem 33 Let M R n+m n+m, where A R n n, A exists, and A π B 0, Let S D CA B If S exists, then 1 M exists if and only if P A + BS π C is regular, P πr A AP πl 0 and S π CA π 0, for some P 1 P{1}; If M exists, then M M 1 M, where AP 1 BS CA + IAP 1 AP 1 BS M 1 S π CP 1 I + BS CA AP 1 S CA AP 1 S S π CP 1 BS A BS M CA π BS π CA π S,

6 H Zhang / Filomat 31:1 017, Proof 1: The Only if part Since M exists, we have that Lemma 1, there exist matrices X and Y over R such that M M X YM By computations, A π B 0, and P A + BS π C, we have M A 0 S π C S 1 M, M 1 P 0 0 S, where 1, are the following invertible matrices, I BS 1 CA I + CA BS, I S C A B I + S CA B, and I + BS 1 1 CA BS CA I I + A, 1 BS C A B S C I From M exists, we have M is group invertible, so is also regular By M 1 dia P, S, it is easy to see that P is regular Let X 1 X1 X Y1 Y, Y X 3 X 4 Y 3 Y 1 1, 4 by the equations M M X and M YM, we have P 0 X1 X A BS 0 S CA π BS π X 3 X 4 CA π S and Y1 Y Y 3 Y 4 From above the two equations, we have P 0 0 S A 0 S π C S PX 1 A BS CA π, 17 PX BS π, S X 3 CA π, S X 4 S, Y 1 P A, Y S 0, Y 3 P S π C, Y 4 S S From 17, we get PX 1 A A A, so PP 1 A A, ie, P πr A 0 By 1, we have AP 1 P A, ie, AP πl 0 Using 19, we get S π CA π 0 The if part Let X 1 P 1 AI A BS CA π, X P 1 BS π, X 3 S CA π, X 4 S and Y 1 AP 1, Y 0, Y 3 S π CA AP 1, Y 4 S Note that, by A π B 0, P πr A AP πl 0 and S π CA π 0, it is easy to verify X 1, X, X 3, X 4 and Y 1, Y, Y 3, Y 4 satisfy the Eqs17 0 and 1 4, respectively That implies M M X YM have solution X, Y, so M exists

7 H Zhang / Filomat 31:1 017, By Lemma 1, we can compute that M Y M Y1 Y Y 3 Y 4 AP 1 0 S π CA AP 1 S M 1 M 1 Y1 Y 1 Y 3 Y 4 I + BS CA BS CA 1 1 1M I AP 1 0 S π CA AP 1 S A BS CA π BS π CA π S Example for Theorem 33: Let Z be the integer ring, and let M be a matrix over Z/6Z, where A, B, C 1 0 4, D Then Let P 1 A 0 4 S 4 0, A π , Sπ, A π B 0, S D CA B , P A + BS π C Therefore P πr A AP πl 0, S π CA π 0 By Theorem, we have M By computations, from Theorem, we can obtain the following corollaries Corollary 34 [13, Theorem31][14, Corollary31], M R n+m n+m, where A R n n is invertible and S D CA 1 B is group invertible, then i M exists if and only if P A + BS π C is invertible; ii If M exists, then M M1 M, where M 3 M 4 M 1 AP 1 A + BS CP 1 A, M AP 1 A + BS CP 1 BS π AP 1 BS, M 3 S π CP 1 A + BS CP 1 A S CP 1 A, M 4 S π CP 1 A + BS CP 1 BS π S CP 1 BS π S π CP 1 BS + S Proof When A is invertible, we have A π P πr P πl 0 Further, PP 1 I P 1 P, this implies P is invertible and P 1 P 1 Whence the corollary is easily proved,

8 H Zhang / Filomat 31:1 017, Corollary 35 [19, Theorem1], M R n+m n+m, where A R n n and S D CA B are group invertible, A π B 0 and S π C 0,then M exists and A M + A BS CA A BS I A BS CA π A BS π S CA S S CA π I Proof Note that S π C 0, P A and P 1 A, the proof immediately follows from Theorem Corollary 36 [0, Theorem5], M R n+m n+m, where A R n n and S D CA B are group invertible, A π B 0, CA π 0 and S π C 0, then M exists and A M + A BS CA A I + BS CA A BS π A BS S CA S I CA BS π Proof Note that CA π 0, the proof immediately follows from Corollary 3 Acknowledgments The authors show great thanks to the editor and referee for their careful reading of the manuscript and valuable comments which greatly improved the readability of the paper References [1] KPS Bhaskara Rao: The theory of generalized inverses over commutative rings, Taylor and Francis 00 [] Wei, Y, Diao, H: On group inverse of singular Toeplitz matrices Linear Algebra Appl 399, [3] Catral, M, Olesky, DD, Driessche, P: Group inverse of matrices with path graphs Electron J Linear Algebra 11, [4] Meyer, CD: The role of the group generalized inverse in the theory of finite Markov chains SIAM Rev 173, [5] Bu, C, Zhang, K: Representations of the Drazin inverse on solution of a class singular differential equations Linear Multilinear Algebra 59, [6] Cao, C, Li, J: Group inverses for matrices over a Bézout domain Electron J Linear Algebra 18, [7] Ben-Israel, A, Greville, TNE: Generalized Inverses: Theory and Applications, nd ed Spinger, New York 003 [8] Cao, C, Li, J: A note on the group inverse of some block matrices over skew fields Appl Math Comput 17, [9] Liu, X, Yang, H: Further results on the group inverses and Drazin inverses of anti-triangular block matrices Applied Math Comput 18, [10] Zhao, J, Hu, Z, Zhang, K, Bu, C: Group inverse for block matrix with t-potent subblock J Appl Math Comput 39, [11] Cao, C, Zhao, C: Group inverse for a class of anti-triangular block matrices over skew fields J Appl Math Comput 40, [1] Ge, Y, Zhang, H, Sheng, Y, Cao, C: Group inverse for two classes of anti-triangular block matrices over rihgt Ore domains J Appl Math Comput 40, [13] Bu, C, Li, M, Zhang, K, Zheng, L: Group inverse for the block matrices with an invertible subblock Appl Math Comput 15, [14] Castro-González, N, Robles, J, Vélez-Cerrada, JY: The group inverse of matrices over a ring Linear Algebra Appl 438, , 013 [15] Campbell, SL: The Drazin inverse and systems of second order linear differential equations Linear Multilinear Algebra 14, [16] Bu, C, Zhang, K, Zhao, J: Some results on the group inverse of the block matrix with a subblock of linear combination or product combination of matrices over shew fields Linear Multilinear Algebra 58, [17] Zhang, K, Bu, C: Group inverses of matrices over right Ore domains Appl Math Comput 18, [18] Liu, X, Wu, L, Benítez, J: On the group inverse of linear combinations of two group invertible matrices Electron J Linear Algebra, [19] Deng, C, Wei, Y: Representations for the Drazin inverse of block-operator matrix with singular Schur complement Linear Algebra Appl 435, [0] Liu, X, Yang, Q, Jin, H: New representation of the group inverse of block matrices J Appl Math [1] Bu, C, Feng, C, Dong, P: A note on computational formulas for the Drazin inverse of certain block matrices J Appl Math Comput 38, [] Sheng, Y, Ge, Y, Zhang, H, Cao, C: Group inverses for a class of block matrices over rings Appl Math Comput 19, [3] Li, B, Bu, C: A note on the group inverse of anti-triangular block matrices over skew filed J Harbin Engineering Univ, 34, [4] Patrício, P, Hartwig, RE: The,, 0 Drazin inverse problem Linear Algebra Appl 437, , 01